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Levart [38]
4 years ago
7

Drag each expression to show whether it is equivalent to 54x + 18, 18(3x – 1), or (6 • 9x) + (6 • 1).

Mathematics
1 answer:
yulyashka [42]4 years ago
4 0

Answer:

See the attachment.

Step-by-step explanation:

In each case, the variable term is 54x, so what makes it be one answer or another is the value of the constant term. The various expressions have one of these constants: +18, -18, or +6.

The two expressions with minus signs go in the "-18" bin.

The two expressions with either +6 or 3(__+2) will go into the "+6" bin.

The remaining two expressions go into the "+18" bin.

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Why can’t one of the names be PNM, or MPN for the plane? But MNP can? What’s the difference, and why?
Savatey [412]

The names PNM and MPN are perfectly valid. The order of the letters doesn't matter. In fact, there are 6 ways to name this plane if we just use the 3 letters. If we include "plane Z", then there are 7 total ways to name the plane.

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3 0
3 years ago
Two corporate baseball teams are scheduled to play a game together. They agree that if both teams attend or if neither team atte
sp2606 [1]

Answer:

2.99% probability that the cost will be paid by only one team

Step-by-step explanation:

The binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Probability that a team plays:

A team plays if it has at most 2 injured players out of 11.

11 players, so n = 11

Each player with a 5% probability of injury, so p = 0.05

Then

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{11,0}.(0.05)^{0}.(0.95)^{11} = 0.5688

P(X = 1) = C_{11,1}.(0.05)^{1}.(0.95)^{10} = 0.3293

P(X = 2) = C_{11,2}.(0.05)^{2}.(0.95)^{9} = 0.0867

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5688 + 0.3293 + 0.0867 = 0.9848

Each team has a 0.9848 probability of showing up to play.

What is the probability that the cost will be paid by only one team?

This happens if one team shows up and the other do not.

2 teams, so n = 2

Each team has a 0.9848 probability of showing up to play, so p = 0.9848.

This probability is P(X = 1).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{2,1}.(0.9848)^{1}.(0.0152)^{1} = 0.0299

2.99% probability that the cost will be paid by only one team

7 0
4 years ago
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